Precalculs

Consider the ellipse 4x^2+49y^2=36
I can figure the vertices is (3,0)(-3,0)
I cannot figure out the foci
Thanks

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  1. 4x^2/36 + 49y^2/36 = 1
    x^2/9 + y^2/(36/49) = 1

    a = 3
    b = 6/7
    c^2 = a^2-b^2 = 9-36/49 = 405/49
    c = 9√5/7

    You found the vertices ok.
    They are at (±a,0)
    The foci are at (±c,0)

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    posted by Steve

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